There is a third subspace connected to an $m \times n$-matrix $A$, namely, the subspace generated by the rows. This is a subspace of $\mathbb{R}^n$, and it may come as a small surprise that it has the same dimension as the column space of $A$, which is a subspace of $\mathbb{R}^m$. This dimension we will call the rank of a matrix.
[Def] The rank of a matrix is defined as the dimension of its column space
$$
\operatorname{rank} A = \dim \operatorname{Col} A
$$
[Q]
For which value(s) of $h$ is the rank of the matrix equal to 3?
$
\begin{pmatrix}
1 & 2 & -h & 1 \\
1 & h+1 & -h-1 & h+1 \\
-2 & -h-3 & h-1 & -h-4 \\
2 & 3h+1 & 1 & 4h+5
\end{pmatrix}
$
[A]
The correct answer is -2, 1. First, to get three zeros in the first column: subtract the first row from the second row AND add the first row two times to the third row AND subtract it two times from the fourth row.
$
A = \begin{pmatrix}
1 & 2 & -h & 1 \\
1 & h+1 & -h-1 & h+1 \\
-2 & -h-3 & h-1 & -h-4 \\
2 & 3h+1 & 1 & 4h+5
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 2 & -h & 1 \\
0 & h-1 & -1 & h \\
0 & 1-h & h-1 & -h-2 \\
0 & 3h-3 & 2h+1 & 4h+3
\end{pmatrix}
$
Two more steps lead to an echelon matrix:
$
A \sim
\begin{pmatrix}
1 & 2 & -h & 1 \\
0 & h-1 & -1 & h \\
0 & 1-h & h-1 & -h-2 \\
0 & 3h-3 & 2h+1 & 4h+3
\end{pmatrix}
\sim
\begin{pmatrix}
1 & 2 & -h & 1 \\
0 & h-1 & -1 & h \\
0 & 0 & -h-2 & -2 \\
0 & 0 & 0 & h-1
\end{pmatrix}
$
Recall that the rank of $A$ is equal to the number of pivots.
From this last matrix we read off that there are four pivots unless if $h = -2$ or $h = 1$.
It is also clear that there are three pivots in case $h = -2$.
For $h = 1$ we further reduce
$
\begin{pmatrix}
1 & 2 & -h & 1 \\
0 & h-1 & -1 & h \\
0 & 0 & -h-2 & -2 \\
0 & 0 & 0 & h-1
\end{pmatrix}
=
\begin{pmatrix}
1 & 2 & -1 & 1 \\
0 & 0 & -1 & 1 \\
0 & 0 & -3 & -2 \\
0 & 0 & 0 & 0
\end{pmatrix}
$
which yields
$
\begin{pmatrix}
1 & 2 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}
$
So for $h = 1$ the rank of $A$ is 3 as well.